chap2

32

Trig

onom

etry

Trigonometry22This chapter at a glanceStage 5.1/5.2/5.3After completing this chapter, you should be able to:�� use trigonometry to find lengths of sides in right-angled triangles where

the given angle is measured in degrees and minutes�� use trigonometry to find sizes of angles measured in degrees and minutes

in right-angled triangles�� solve practical problems that involve finding the lengths of sides and sizes

of angles in right-angled triangles, where the angles are measured in degrees and minutes

�� solve practical problems that involve angles of elevation and depression, where the angles are measured in degrees and minutes

�� express the tangent ratio in terms of the sine and cosine ratios and use this result to solve problems

�� state the complementary results for the sine and cosine ratios and use these results to solve problems

�� state the exact values and use them to solve problems�� draw diagrams showing the compass bearing of one point from another

point�� draw diagrams showing the true bearing of one point from another point�� find the bearing of A from B given the bearing of B from A�� solve practical trigonometry problems involving bearings.

Mathscape 10 ext. - Ch02 Page 32 Thursday, October 13, 2005 2:38 PM

Chapter 2 : Trigonometry 33

The trigonometric ratios are the ratios of pairs of sides in right-angled triangles. These ratios in relation to an angle θ are the sine ratio (sin θ), the cosine ratio (cos θ) and the tangent ratio (tan θ).

NOTE: The abbreviations SOH CAH TOA can be used to help you remember these definitions.

Example 1Find values for sin θ, cos θ and tan θ in the given triangle.

Solutions

sin θ = cos θ = tan θ =

= = =

Example 2Find value of x, given that sin θ = .

Solution

sin θ =

=

× 35 × 35

∴ x = 15

1 Name the opposite side, the adjacent side and the hypotenuse in each triangle.a b c

2.1 The trigonometric ratios

The definitions of the trigonometric ratios are:

� ��

��

adjacent

hypotenuse

oppo

site

θ

sin θ oppositehypotenuse-----------------------------= cos θ adjacent

hypotenuse-----------------------------=

tan θ oppositeadjacent----------------------=

EG +S

12

135

θ

oppositehypotenuse--------------------------- adjacent

hypotenuse--------------------------- opposite

adjacent--------------------

513------ 12

13------ 5

12------

x35

θEG +S

37---

37---

x35------ 3

7---

Exercise 2.1

QP

R

θG

E Fθ

Y

ZX

θ

Mathscape 10 ext. - Ch02 Page 33 Friday, September 23, 2005 6:47 AM

Mathscape 10 extens i on34

2 For each of the following triangles, state as a fraction the value of:i sin θ ii cos θ iii tan θa b c

d e f

■ Consolidation

3 Name the angle that has a:

a tangent of b sine of

c cosine of d tangent of

e cosine of f sine of

4 Find, without simplifying, the value of:a sin ∠PRQ b tan ∠PSQc cos ∠SPQ d sin ∠QPRe tan ∠QPR f cos ∠PRQg sin ∠PSQ h tan ∠SPQi cos ∠QPR j sin ∠SPQk tan ∠PRQ l cos ∠PSQ

5 Find the value of the pronumeral in each triangle.a b c

tan θ = sin θ = cos θ =

d e f

tan θ = cos θ = sin θ =

5

4

3

θ15

178

θ

13

512

θ

21

20 29

θ

24 7

25θ

1665

63θ

A

CB

5328

45θ

α 4528------ 28

53------

4553------ 28

45------

2853------ 45

53------

P

SRQ

25

12

1715

8

x 6

θy15

θ

p32

θ

12--- 3

5--- 5

8---

c

21

θe

55

θn

20

θ

79--- 11

12------ 4

7---

Mathscape 10 ext. - Ch02 4 Friday, September 23, 2005 6:47 AM


6 Find the values for sin θ, cos θ, tan θin this trapezium.

7 EFGH is a rhombus with diagonals EG and FH of length 32 cmand 126 cm respectively.a Find the side length of the rhombus.b Find values for sin θ, cos θ, tan θ.

8 In the isosceles triangle XYZ, W is a point on XZ such that YW ⊥ XZ. If XY = YZ = 89 mm and XZ = 78 mm:a find the length of YWb find values for sin α, cos α and tan α.

■ Further applications

9 a If , find values for sin θ, cos θ.

b If , find values for cos θ, tan θ.

c If , find values for tan θ, sin θ.

10 If , find expressions for sin θ, cos θ.

Angles are measured in degrees, minutes and seconds (1 degree = 60 minutes, i.e. 1° = 60′ and 1 minute = 60 seconds, i.e. 1′ = 60″). The degrees, minutes and seconds key on your calculator should look like either or .

When rounding off angles correct to the nearest degree, angles with less than 30 minutes are rounded down, while angles with 30 minutes or more are rounded up. Similarly, when rounding off angles correct to the nearest minute, angles with less than 30 seconds are rounded down, while angles with 30 seconds or more are rounded up.

20

85 36

97θ

E F

GH

θ

Y

ZX W

89 mm

78 mm

α

tan θ 2021------=

sin θ 2425------=

cos θ 5573------=

tan θ xy--=

2.2 Degrees and minutes

DMS ° ′ ″



■ Evaluating trigonometric expressions

A calculator can be used to find the value of a trigonometric expression. The order in which you press the keys will vary between calculator models.

■ Finding an angle

The inverse key , shift key , or second function key , can be used to undo the process of finding the sine, cosine or tangent of an angle, and hence to find the size of an angle.

Example 1Round off each angle correct to the nearest degree.

a 15°26′ b 37°43′ c 75°30′

Solutionsa 26′ is less than 30′, so we round down: 15°26′ � 15°.b 43′ is greater than 30′, so we round up: 37°43′ � 38°.c 30′ is halfway, so we round up: 75°30′ � 76°.

Example 2Round off each angle correct to the nearest minute.

a 21°38′19″ b 52°13′49″ c 5°56′30″

Solutionsa 19″ is less than 30″, so we round down: 21°38′19″ � 21°38′.b 49″ is greater than 30″, so we round up: 52°13′49″ � 52°14′.c 30″ is halfway, so we round up: 5°56′30″ � 5°57′.

Example 3Evaluate each expression, correct to 2 decimal places.

a sin 23°16′ b 19 cos 8°39′ c

To evaluate a trigonometric expression:�� press the appropriate trigonometric ratio key, then enter the angle.

INV shift 2nd F

To find the size of an angle given either a fraction or a decimal:�� press the , or key followed by the appropriate trigonometric

function key�� enter the fraction or decimal into the calculator, then press �� round off the angle as required.

INV shift 2nd F

=

EG +S

EG +S

EG +S 31

tan 12°42′-------------------------



Solutions

Example 4

Evaluate correct to 3 significant figures.

SolutionCalculator steps: 42 9 8 54 Calculator readout: 0.780774459Answer: 0.781

Example 5Find θ, correct to the nearest minute.

a tan θ = 1.3759 b sin θ = 0.1382

Solutions

1 Round off each angle, correct to the nearest degree.a 19°56′ b 36°15′ c 20°30′ d 71°24′e 62°50′ f 84°9′ g 109°35′ h 137°52′

2 Round off each angle, correct to the nearest minute.a 7°51′35″ b 22°6′13″ c 14°25′30″ d 76°32′49″e 50°16′8″ f 68°4′30″ g 119°12′37″ h 164°2′10″

■ Consolidation

3 Evaluate these trigonometric expressions, correct to 2 decimal places.a cos 37°8′ b tan 6°29′ c sin 21°54′ d cos 43°35′e 9 sin 57°18′ f 25 cos 8°26′ g 16.3 tan 49°50′ h 45.7 sin 20°32′

i j k l

Calculator steps Calculator readout Answer

a 23 16 0.395011105 0.40

b 19 8 39 18.7838847 18.78

c 31 12 42 137.5578485 137.56

Calculator steps Calculator readout Answer

a 1.3759 53°59′25.65″ 53°59′b 0.1382 7°56′37.32″ 7°57′

sin ° ′ ″ ° ′ ″ =

cos ° ′ ″ ° ′ ″ =÷ tan ° ′ ″ ° ′ ″ =

EG +S

cos 42°9′tan 8° sin 54°+--------------------------------------

cos ° ′ ″ ° ′ ″ ÷ ( tan + sin ) =

EG +S

shift tan = ° ′ ″shift sin = ° ′ ″

Exercise 2.2

13tan 25°11′------------------------- 29.8

sin 5°27′--------------------- 1

cos 44°20′------------------------- 1

tan 72°55′-------------------------



4 Evaluate each expression, correct to 4 significant figures.

a b c

5 Evaluate each expression, correct to 4 significant figures.

a b

6 Find the acute angle θ, correct to the nearest minute.a tan θ = 0.3675 b cos θ = 0.7173 c sin θ = 0.1247d cos θ = 0.2662 e sin θ = 0.9641 f tan θ = 0.0899g sin θ = 0.5863 h tan θ = 1.4085 i cos θ = 0.9999j tan θ = 2.7891 k cos θ = 0.6124 l sin θ = 0.4207

7 Find the acute angle θ, correct to the nearest minute.a tan θ = b sin θ = c cos θ = d sin θ =


8 a If cos θ = 0.7923, find tan θ and sin θ, correct to 4 decimal places.b If tan θ = 4.0672, find sin θ and cos θ, correct to 4 decimal places.

9 Find the acute angle α, correct to the nearest minute.

a 3 sin α + 5 = 6 b c

Pythagoras’ theorem is used to find the length of a side in a right-angled triangle when the lengths of the other two sides are known. Trigonometry is used to find the length of a side when the length of one other side and the size of one angle are known.

■ Finding the length of a short side

NOTE: In those questions involving the tangent ratio where the pronumeral would be in the denominator, it is often easier to find the other acute angle in the triangle and use it to find the required side length. By using the other angle, the pronumeral should then be in the numerator.

9 sin 53°28′8 cos 8°14′----------------------------- 17 tan 73°45′

21 sin 12°29′-------------------------------- 23 cos 15°34′

5 tan 9°56′---------------------------------

sin 40°18′ tan 26°11′+cos 47°19′

-------------------------------------------------------- cos 4°37′sin 52°25′ tan 9°14′–----------------------------------------------------

25--- 9

10------ 3

4--- 6

11------

7 tan α3

----------------- 4= 2cos α------------- 9=

2.3 Finding the length of a side

To find the length of the opposite or adjacent sides:�� determine which ratio is to be used�� write down a trigonometric equation�� multiply both sides by the denominator�� evaluate using a calculator.



■ Finding the length of the hypotenuse

Example 1Find the value of the pronumeral in each of the following, correct to 1 decimal place.

a b

Solutionsa tan 61°8′ = b cos 17°28′ =

× 39 × 39 × 52 × 52y = 39 × tan 61°8′ h = 52 × cos 17°28′

= 70.74562525 = 49.60237011= 70.7 (1 decimal place) = 49.6 (1 decimal place)

Example 2Find the length of the hypotenuse, correct to 1 decimal place.

Solutionsin 23°46′ =

=

× 11 × 11

p =

= 27.2944099

∴ The length of the hypotenuse is 27.3 cm (1 decimal place).

To find the length of the hypotenuse:�� determine whether the sine or cosine ratio is to be used�� write down a trigonometric equation�� take the reciprocal of both sides�� multiply both sides by the denominator under the pronumeral�� evaluate using a calculator.

EG +S

39 cm

y cm

61°8'

52 cm

h cm17°28'

y39------ h

52------

11 cmp cm

23°46'

EG +S

11p

------

1sin 23°46′------------------------ p

11------

11sin 23°46′------------------------



1 Find the value of the pronumeral in each of these, correct to 1 decimal place. All lengths are in centimetres.a b c

d e f

g h i

j k l

m n o

2 Find the value of the pronumeral in each triangle, correct to 2 decimal places. All lengths are in metres.a b c

Exercise 2.3

p

3010°28'

m

19

17°32'

x35°9'

43

y

64°6'54

t

82°14'

1253°41'

a

72

g34°11'

61

b

29°40'23 k

47

55°16'

z

22°57'

78.9c

68°25'

16.4

w

55.1

9°28'

e

53°10'40.6

n77°51'

84.5u

40°8'104.3

x

14

15°24'

t

23

20°39'e

36

35°58'



d e f

■ Consolidation

3 Draw a diagram and mark on it all of the given information, to answer each of the following.a In ∆XYZ, ∠Y = 90°, ∠X = 54°17′ and XZ = 25 cm. Find the length of XY, correct to

1 decimal place.b In ∆PQR, ∠R = 90°, ∠Q = 21°4′ and QR = 43 m. Find the length of PR, correct to

4 significant figures.c In ∆LMN, ∠L = 90°, ∠N = 16°45′ and MN = 37 cm. Find the length of LM, correct to

the nearest millimetre.d In ∆ABC, ∠A = 90°, ∠B = 47°50′ and AC = 85 cm. Find the length of BC, correct to

3 significant figures.

4 A cruise ship is anchored 300 m from the base of a vertical cliff. A passenger stood on the bow of theship and looked up to the top of the cliff at an angle of 68°23′. Calculate the height of the cliff, correct to the nearest metre.

5 A wire of length 7.3 m attached to the top of a flagpole is inclinedto the ground at an angle of 61°49′. Find the height of the flagpole,correct to the nearest tenth of a metre.

6 A ladder of length 13.4 m leans against a wall and makes an angle of 56°12′ with the ground. How far is the foot of the ladderfrom the base of the wall? Answer correct to the nearest tenth ofa metre.

m

69°21'

56.3

c

7°40'

19.8

v

36°53'

25.4

h m

300 m68°23'

7.3 m

61°49'

h m

56°12'

13.4 m

x cm



7 A school building casts a shadow 32 m long on theplayground when the sun has an altitude of 29°44′.Calculate the height of the building, correct to the nearest metre.

8 The window awning outside a butcher shopis attached to the front wall at an angle of52°35′. Find the width of the awning given that it reaches 155 cm out from the wall.Answer correct to the nearest centimetre.

9 Elicia is flying a kite at the local park. She is 1.4 m tall and thehand holding the string is just above eye level. The string is23 m long and makes an angle of 64°7′ with the vertical.How high is the kite above the ground? Answer correct to thenearest tenth of a metre.

10 Find the height, h cm, of this trapezium, correct to 1 decimal place.

11 In the parallelogram PQRS, PS = 12 cm, PQ = 26 cmand ∠PQR = 61°25′.a Find PT, the height of the parallelogram, correct

to 1 decimal place.b Hence, calculate the area of the parallelogram.

29°44'32 m

h m

155 cm

52° 35'

w cm

h m

64°7'

23 m

h cm

143°8'15 cm

21 cm

P

ST R

Q

12 cm

26 cm

61°25'



12 In the semicircle, centre O, OF = 12.5 cm and ∠FEG = 73°44′.a Write down the length of the diameter EG. b Find the length of the chord FG, correct to

the nearest centimetre.

13 The diagonal in a rectangle makes an angle of 26°39′ with the length. If the rectangle has a length of 15 cm, find the following correct to 1 decimal place:a the width of the rectangle b the length of a diagonal

14 A straight road rises at an angle of 19°16′ to the horizontal and has a vertical rise of 115 m. Find the length of the road, correct to the nearest metre.

15 When the sun has an altitude of 67°42′, a pine tree casts a shadow 8.3 m long. Find the height of the tree. Answer correct to the nearest tenth of a metre.

16 A wedge-shaped ramp is set up for a car stunt on a movie set. The ramp has an angle of inclination of 21°37′ and a vertical rise of 2.45 m. Find the length of the ramp, correct to the nearest centimetre.

17 A surveyor sights the top of a mountain from a distance of 1500 m. The angle of elevation of the top of the mountain is 51°17′. Find the height of the mountain, correct to the nearest metre.

18 A wall casts a shadow of length 24.6 m along the ground. If the angle of elevation of the sun is 18°23′ from the end of the shadow, find the height of the wall, correct to the nearest centimetre.

19 From a window on the fifth floor of an office tower, a worker observes a ferry that has just moored across the street at Circular Quay. The window is 20 m above the street and the angle of depression of the ferry from the window is 21°40′. How far is the ferry from the base of the building? Answer correct to 3 significant figures.

20 From the top of the control tower at the airport, the angle of depression of a plane on the tarmac is 39°25′. If the height of the control tower is 47 m, find the distance between the plane and the tower, correct to the nearest metre.

21 A plane takes off at a speed of 300 km/h and rises at an angle of 36°49′. After 3 minutes the plane levels off and continues flying at a constant altitude.a Calculate the distance in metres that the plane flies through the air, before reaching its

cruising altitude.b Find the cruising altitude of the plane in metres, correct to 1 significant figure.

22 In the isosceles ∆PQR, PQ = PR and S is a point on QR such that PS ⊥ QR.a If ∠PQR = 74°53′ and PQ = 85 mm, find the length of QR, correct to the nearest

millimetre.b If ∠PQR = 62°35′ and QR = 126 mm, find the length of PQ, correct to 1 decimal place.

F

GEO

73°44'

12.5 cm



c If ∠QPR = 104°28′ and PS = 93 mm, find the length of QR, correct to 3 significant figures.

d If ∠QPR = 81°36′ and QR = 138 mm, find the length of PS, correct to 2 significant figures.

23 A cone has a diameter of 31.6 cm and the angle between the vertical axis and the slant height is 15°52′. Find the perpendicular height of the cone, correct to 4 significant figures.


24 In the kite ABCD, AB = BC = 15 cm, AD = DC, ∠ACB = 38°10′ and ∠ACD = 61°40′.a Find the length of the diagonal AC, correct to

1 decimal place.b Find the length of the diagonal BD, correct to

1 decimal place.c Hence, calculate the area of the kite.

25 In the rectangular prism shown, FG = 6 cm, HG = 8 cm and ∠BHF = 60°.a Find the length of FH.b Hence, find the length of the diagonal BH.

26 Two buildings of different heights stand on opposite sides of the street, 35 m apart. A man standing on the roof of the shorter building measures the angle of elevationof the top of the taller building as 54°57′. The shorter building is 75 m tall. Calculate the height of the tallerbuilding, correct to the nearest metre.

27 a Find the length of AD and DC, correct to 1 decimal place.

b Hence, find the length of AC.

B

A

D

C

15 cm38°10'

61°40'

A B

F

G

E

H

D C

60°

8 cm6 cm

54°57'h m

75 m

35 m

B

AD C

68° 47°

12 cm



28 a Find the length of QS, correct to 1 decimal place.

b Hence, find the length of QR, correct to 1 decimal place.

29 From the top of a lighthouse situated 100 m abovesea level, the angles of depression of two boats at seaare 27°39′ and 52°46′ respectively. Calculate thedistance between the boats, correct to the nearestmetre.

Trigonometry can also be used to find sizes of angles in triangles. In exercise 2.2 we used the

inverse key , shift key , or second function key , followed by the ,

or key to find the size of an angle given a fraction or decimal value. When we find an angle, we are actually using the inverse trigonometric functions—sin−1x, cos−1x and tan−1x. The notation sin−1x, read as the inverse sine of x, means ‘the angle whose sine is x’. The expressions cos−1x and tan−1x have similar meanings. The inverse trigonometric functions ‘undo’ the basic trigonometric functions of the sine, cosine and tangent ratios, and hence give the size of the angle.

SRP

Q

70°14°

18 cm

27°39'

52°46'

x m

100 m

Swing

The supporting bar of a playground swing is 4 m above the ground and the swing itself is 3.5 m long. A man places his child on the swing and pulls it back until the seat is 1.7 m above the ground. Through what angle will the child swing? Give your answer to the nearest degree.

3.5 m

4 m

1.7 m

TRY THIS

2.4 Finding the size of an angle

INV shift 2nd F sin cos

tan

The inverse trigonometric functions are used to find the size of an angle, where:�� sin−1x means ‘the angle whose sine is x’�� cos−1x means ‘the angle whose cosine is x’�� tan−1x means ‘the angle whose tangent is x’.



NOTE: If the side lengths are decimals you may need to use the division and grouping symbols keys on the calculator rather than the fraction key.

Example 1Find the size of the angle θ, correct to the nearest minute.

a b

Solutions

a cos θ =

θ = cos−1

Press 9 22

θ = 65°51′8.14″= 65°51′ (to the nearest minute)

In this exercise, find all angles correct to the nearest minute, unless otherwise stated.1 Find the size of the angle θ. All lengths are in metres.

a b c

d e f

To find the size of an angle:�� press the , or key followed by the appropriate trigonometric

function key�� enter the given fraction or decimal, then press �� round off the angle as required.

INV shift 2nd F

=

EG +S

9

22θ

17

4θ

b tan θ =

θ = tan−1

Press 17 4

θ = 76°45′34.13″= 76°46′ (to the nearest minute)

174

------

174

------⎝ ⎠⎛ ⎞

shift tan abc--- = ° ′ ″

922------

922------⎝ ⎠

⎛ ⎞

shift cos abc--- = ° ′ ″

Exercise 2.4

815

θ

11

13θ

9

20θ

3

14

θ

21

25θ

18

17θ



g h i

j k l

2 Find the size of the angle α. All lengths are in millimetres.a b c

d e f

g h i

■ Consolidation

3 Find the size of the angle φ. All lengths are in centimetres.a b c

1637

θ

27

70θ

10

49

θ

26

41

θ

95

37

θ42

53θ

7.2

10.9α

9.3

15.7α

17.8

14.1

α

19.5

5.4α

22.6

8.1α

20.3

30.4

α

72.2

8.5

α81.9

34.6α

126.4

91.7α

3p

2p

φ8x

13x

φ

11c 9c

φ



4 Draw a diagram and mark on it all of the given information, to answer each of the following.a In ∆CDE, ∠D = 90°, CD = 9 cm and DE = 17 cm. Find ∠C.b In ∆OPQ, ∠Q = 90°, OQ = 13 mm and OP = 14.8 mm. Find ∠P.c In ∆IJK, ∠I = 90°, IK = 12.7 m and JK = 15.9 m. Find ∠K.

5 A cone has a diameter of 36 cm and a perpendicular height of 23 cm. Find the size of the semi-vertical angle θ.

6 A boat is to be launched down a 25 m ramp with a vertical drop of 8 m. At what angle is the ramp inclined to the horizontal?

7 A girl observes a bird’s nest at the top of a tree. She stands 7.5 m from the base of the tree, which is 16.5 m tall. At what angle must the girl look up in order to see the nest?(Ignore the girl’s height.)

8 A chimney stack 49.6 m tall casts a shadow 77.2 m long on theground. Calculate the angle at which the sun’s rays strike theground.

9 A rectangle has a length of 17 cm and diagonals of length 21 cm. Find the angle between the width and a diagonal.

36 cm

23 c

m

θ

8 m

25 m

θ

7.5 m

16.5 m

θ

77.2 m

θ

49.6 m

21 cm

17 cm

θ



10 The anchor of a boat was dropped 45 m to the sea floor.During the next hour, the boat drifted 16 m from thisspot. Find the angle between the anchor line and thesurface of the water.

11 While trying to swim across a river, Brad was sweptdownstream by the current. The river is 42 m wide, but Brad had to swim 60 m to get across. At what anglewas he dragged downstream?

12 a A ladder of length 6 m leans against a wall. The foot of the ladder is 1.3 m from the base of the wall. Find the angle formed between the ladder and the ground.

b A ladder of length 7.3 m reaches 5.95 m up the side of a wall. What angle does the ladder make with the wall?

13 A straight stretch of railway track has a gradient of . Calculate the angle at which the track rises.

14 A grass ski run falls 125 m vertically over a sloping run of 160 m. Find the angle at which the ski run is inclined to the horizontal.

15 In the isosceles ∆RST, RS = ST and U is a point on RT such that SU ⊥ RT.a If SU = 17 cm and RT = 26 cm, find the size of ∠SRT.b If ST = 29 cm and RT = 43 cm, find the size of ∠STR.c If RS = 33 cm and SU = 21 cm, find the size of ∠RST.d If RT = 50 cm and SU = 37 cm, find the size of ∠RST.

16 A pendulum of length 85 cm swings through a horizontal distance of 60 cm before it stops and swings back again. Find the vertical angle through which the pendulum swings.

17 A cone has a diameter of 46 cm and a slant height of 63 cm. Find the angle between the slant height and the base of the cone.


18 An observer on the ground sees a plane flying directly overhead at a speed of 360 km/h, and at an altitude of 5000 m. At what angle would the observer need to look up in order to see the plane 5 minutes later?

19 In ∆LMN, ∠M = 90°, MN = 2LM and X is the midpoint of LM. Find the size of ∠LNX.

20 WXYZ is a quadrilateral in which ∠X = 90°, ∠Z = 90°, WX = 15 mm, XY = 20 mm, YZ = 24 mm and WZ = 7 mm. Find the size of ∠W.

16 m

45 m

θ

60 m42 mθ

415------



21 Find the angle θ in each of the following.a b

c d

The tangent ratio can be expressed as the quotient of the sine and cosine ratios. For any angle θ:

Proof: In ∆ABC, let ∠CAB = θ.

∴

Now, =

=

= tan θ

∴ tan θ =

78101

130°

A

BCDθ

30

25

45W X

Y

Z

θ

40°54'

2518

M

L K Nθ

13

5 16

B

CDA

θ

High flying

An airliner flying at a height of 8000 m needs to descend to a height of 200 m when it is 5 km from the airport. If its angle of descent is 3º, how far from the airport should the pilot begin his descent?

TRY THIS

2.5 The tangent ratio

tan θ sin θcos θ------------=

B

C A

a

b

c

θ

sin θ ac--- cos θ, b

c--- and tan θ a

b---= = =

sin θcos θ------------

ac---

bc------

× c

× c

ab---

sin θcos θ------------



Example 1a Find the value of tan θ, where θ is an acute angle, if sin θ = 0.36 and cos θ = 0.4.b Hence, find θ correct to the nearest minute.

Solutions

a tan θ = b tan θ = 0.9

=θ = tan−1 0.9

= 0.9

= 41°59′ (to the nearest minute)

Example 2Express each equation in terms of tan θ, where θ is an acute angle, then solve for θ.Answer correct to the nearest minute.

a sin θ = 4 cos θ b 5.2 sin θ = 2.3 cos θ

Solutionsa sin θ = 4 cos θ

÷ cos θ ÷ cos θ

= 4

tan θ = 4∴ θ = tan−1 4

= 75°58′(to the nearest minute)

1 Verify each of the following by using a calculator.

a b c

2 Find values for sin θ, cos θ and tan θ, then verify that in each triangle.

a b c

EG +S

sin θcos θ-------------

0.360.4

----------

EG +S

b 5.2 sin θ = 2.3 cos θ÷ 5.2 cos θ = ÷ 5.2 cos θ

=

tan θ =

∴ θ = tan−1

= 23°52′(to the nearest minute)

sin θcos θ------------ 2.3

5.2-------

2.35.2-------

2.35.2-------⎝ ⎠

⎛ ⎞

sin θcos θ------------

Exercise 2.5

sin 40°cos 40°----------------- tan 40°= sin 17°38′

cos 17°38′------------------------- tan 17°38′= sin 115°

cos 115°-------------------- tan 115°=

sin θcos θ------------ tan θ=

3

4 5θ

13

12

5

θ 12

37

35

θ



■ Consolidation

3 Find the value of tan θ in each of the following, where θ is an acute angle. Hence, find the size of the angle θ, correct to the nearest minute.

a sin θ = and cos θ = b sin θ = 0.5466 and cos θ = 0.8374

c cos θ = 0.6118 and sin θ = 0.7910 d and

4 If , show that:

a sin θ = cos θ tan θ b

5 a If cos θ = and tan θ = , find the value of sin θ.

b If tan θ = and cos θ = , find the value of sin θ.

c If sin θ = and tan θ = , find the value of cos θ.

d If tan θ = and sin θ = , find the value of cos θ.

6 Solve the equation sin θ = cos θ, where θ is an acute angle.

7 In the equations below, θ is an acute angle. Express each equation in terms of tan θ, then solve for θ, correct to the nearest minute.a sin θ = 2 cos θ b cos θ = 3 sin θc 5 sin θ = 8 cos θ d 13 cos θ = 7 sin θe 6.4 cos θ = 3.9 sin θ f 21.5 sin θ = 16.1 cos θ


8 In the equations below, θ is an acute angle. Express each equation in terms of tan θ, then solve for θ, correct to the nearest minute.

a b c

9 Prove each of the following identities.

a b sin A cos A tan A = sin2 A

c d

e f

817------ 15

17------

cos θ 23

-------= sin θ 73

-------=

tan θ sin θcos θ------------=

cos θ sin θtan θ------------=

725------ 24

7------

940------ 40

41------

2029------ 20

21------

6316------ 63

65------

8sin θ------------ 1

cos θ------------= sin θ

3------------ cos θ

4------------= 5

cos θ------------ 2

sin θ------------=

sin2 Acos2 A---------------- tan2 A=

sin A cos Atan A

---------------------------- cos2 A= sin Acos A tan A---------------------------- 1=

tan A cos Asin A

---------------------------- 1= sin2 A sin A cos A+cos2 A sin A cos A+--------------------------------------------------- tan A=



In any right-angled triangle, the sine of an acute angle is equal to the cosine of its complement, and the cosine of an acute angle is equal to the sine of its complement. That is,

Proof: In ∆ABC, ∠C = 90°, ∠A = θ and ∠B = 90° − θ.

Now, and , ∴ sin θ = cos (90° − θ)

Also, and , ∴ cos θ = sin (90° − θ)

Example 1Find the value of x in each of the following.

a cos x° = sin 70° b sin x° = cos 55°

Solutionsa cos x° = sin 70° b sin x° = cos 55°

cos x° = cos 20° sin x° = sin 35°∴ x = 20 ∴ x = 35

Example 2Find the value of x in each of these.

a cos (x − 20)° = sin 40° b sin (2x)° = cos 10°

Solutionsa cos (x − 20)° = sin 40° b sin (2x)° = cos 10°

cos (x − 20)° = cos 50° sin (2x)° = sin 80°∴ x − 20 = 50 ∴ 2x = 80

∴ x = 70 ∴ x = 40

Example 3

Simplify without using a calculator.

Solution

=

= 3

2.6 The complementary results

sin θ = cos (90° − θ) and cos θ = sin (90° − θ)

B

C A

c

b

a

90° – θ

θ

sin θ ac---= cos 90° θ–( ) a

c---=

cos θ bc---= sin 90° θ–( ) b

c---=

EG +S

EG +S

EG +S 3 sin 75°

cos 15°---------------------

3 sin 75°cos 15°

--------------------- 3 sin 75°sin 75°

---------------------



1 a Find values for each of the following, using the given triangle.i sin θ ii cos θ

iii sin (90° − θ) iv cos (90° − θ)b Use your results from part a to complete the following statements.

i sin θ = .......... ii cos θ = ..........

2 Verify the following results by using a calculator.a sin 60° = cos 30° b cos 60° = sin 30° c cos 20° = sin 70°d sin 20° = cos 70° e sin 35° = cos 55° f cos 55° = sin 35°

3 a Simplify 27°39′ + 62°21′.b Verify that sin 27°39′ = cos 62°21′.c Would sin 62°21′ = cos 27°39′? Why?

4 Complete the following statements.a If cos 60° = 0.5, then sin ____ = 0.5b If sin 40° = 0.643, then cos ____ = 0.643c If cos 15° = 0.966, then ____ 75° = 0.966d If sin 72° = 0.951, then ____ 18° = 0.951

■ Consolidation

5 Find the value of x in each of these.a sin 50° = cos x° b cos 35° = sin x° c sin x° = cos 65°d cos x° = sin 12° e cos x° = sin 48° f sin 53° = cos x°g sin 17° = cos x° h sin x° = cos 84° i cos x° = sin 5°j cos 64° = sin x° k sin 71° = cos x° l sin x° = cos 29°

6 Simplify without using a calculator:

a b c d

7 Solve each of these equations.a sin (x + 30)° = cos 20° b sin 35° = cos (x + 10)°c cos 30° = sin (x − 23)° d cos (x − 16)° = sin 26°e sin (2x)° = cos 36° f cos 42° = sin (3x)°

g = cos 30° h = sin 40°

i cos (2x + 56)° = sin 14° j sin x° = cos x°k cos (2x)° = sin x° l sin (x + 18)° = cos (x − 8)°

8 a If cos 75° = 0.26, find the approximate value of cos 75° + sin 15°.b If sin 4° = 0.07, find the approximate value of 3 cos 86°.

Exercise 2.6

b

ca

(90° – )

θ

θ

sin 10°cos 80°----------------- cos 74°

sin 16°----------------- 2 sin 57°

cos 33°--------------------- 3 cos 29°

4 sin 61°----------------------

sin x4---⎝ ⎠

⎛ ⎞ ° cos x2---⎝ ⎠

⎛ ⎞ °




9 Simplify the following expressions.

a b c d

e sin θ cos (90° − θ) f cos θ sin (90° − θ) g tan θ sin (90° − θ)

h i sin (90° − θ) cos (90° − θ) tan (90° − θ)

10 Prove the following identities.

a b

c d

The triangles below can be used to derive exact values for the trigonometric ratios involving angles of 30°, 45° and 60°. This means that the answers to many questions can now be given in exact form as fractions or as surds, rather than as the decimal approximations that are usually obtained by using a calculator.

A right-angled isosceles triangle is used to find the exact values of the trigonometric ratios involving angles of 45°.

In ∆PQR, ∠Q = 90°, PQ = QR = 1 unit and ∠P = ∠R = 45°. PR = units (by Pythagoras’ theorem).

An equilateral triangle is used to find the exact values of the trigonometric ratios involving angles of 30° and 60°.

In ∆PQR, PQ = QR = RS = 2 units and ∠P = ∠Q = ∠R = 60°. Construct QS ⊥ PR. Now QS bisects both PR and ∠PQR. Therefore, PS = SQ = 1 unit and ∠PQS = ∠RQS = 30°. QS = (by Pythagoras’ theorem).

The exact values for the trigonometricratios are summarised in the table.

cos 90° θ–( )sin θ

-------------------------------- cos 90° θ–( )cos θ

-------------------------------- cos θsin 90° θ–( )------------------------------- sin θ

sin 90° θ–( )-------------------------------

cos 90° θ–( )tan θ

--------------------------------

sin θ cos 90° θ–( )cos θ sin 90° θ–( )---------------------------------------------- tan2 θ= sin θ sin 90° θ–( )

cos θ cos 90° θ–( )----------------------------------------------- 1=

cos θ tan θ cos 90° θ–( ) sin θ= sin θ sin 90° θ–( )tan θ

--------------------------------------------- cos θ=

2.7 The exact values

P

Q R

1

1

45°

45°

2

2

RS

Q

P

3

30° 30°

60°60°1 1

223

θ 30° 45° 60°

sin θ

cos θ

tan θ 1

12---

1

2------- 3

2-------

32

-------1

2------- 1

2---

1

3------- 3



Example 1Evaluate each expression without using a calculator.

a tan 30° sin 60° b tan2 60° − sin2 60°

Solutionsa tan 30° sin 60° b tan2 60° − sin2 60°

=

=

Example 2Find the exact value of each expression. Give the answers in simplest surd form with a rational denominator.

a cos 30° + sin 45° b

Solutionsa cos 30° + sin 45°

=

=

=

=

Example 3Solve the equation for θ, where 0° < θ < 90°.

Solution

tan θ =

θ = tan−1

= 30°

EG +S

= (tan 60°)2 − (sin 60°)2

=

=

= 2

3( )2 32

-------⎝ ⎠⎛ ⎞

2

–

3 34---–

14---

1

3------- 3

2-------×

12---

EG +S

sin 45°tan 60°-----------------

b

=

=

=

sin 45°tan 60°-----------------

1

2-------

3-------

×

×

2

2

1

6-------

××

66

66

-------

32

------- 1

2-------+

32

-------1

2-------⎝ ⎠

⎛ ⎞+××

22

32

------- 22

-------+

3 2+2

--------------------

EG +S tan θ 1

3-------=

1

3-------

1

3-------⎝ ⎠

⎛ ⎞



Example 4Find the value of x.

Solution

tan 30° =

=

× ×∴ x = 4

1 Write down the exact value of each expression.a sin 30° b cos 30° c tan 30°d sin 45° e cos 45° f tan 45°g sin 60° h cos 60° i tan 60°

2 Find the exact value of:a tan2 45° b sin2 30° c tan2 60°d cos2 45° e tan2 30° f cos2 30°

3 Find the exact value of each expression. Give your answers in simplest surd form where necessary, with a rational denominator.a sin 30° cos 60° b sin 45° cos 45° c tan 60° tan 30°d sin 60° cos 30° e cos 30° sin 45° f cos 45° tan 60°g sin 45° tan 30° h cos 45° cos 60° i tan 30° cos 30°j sin 30° cos 45° sin 45° k tan 60° cos 30° tan 45° l sin 60° tan 30° cos 30°

■ Consolidation

4 Find the exact value of each expression. Give your answers in simplest surd form where necessary, with a rational denominator.a sin 30° + cos 60° b tan 60° + tan 30° c sin 45° + cos 45°d sin 60° + cos 30° e cos 45° + sin 30° f sin 60° + sin 45°g cos 30° + tan 60° h tan 45° + sin 60° i tan 45° + sin 30°

5 Find the exact value of each expression. Give your answers in simplest surd form where necessary, with a rational denominator.a sin 30° + cos 30° + tan 30° b tan 30° + cos 45° + sin 60°c tan 45° + tan 30° + tan 60° d tan 60° + sin 45° + cos 60°

6 Find the exact value of each expression.a 2 sin2 45° − 1 b 1 − 2 sin2 60° c tan2 60° − tan2 45°d sin2 60° − sin2 30° e 4(sin2 45° + cos2 45°) f tan2 60° − 2 cos2 30°

EG +S

30°

x cm

4 3 cmx

4 3----------

1

3------- x

4 3----------

4 3 4 3

Exercise 2.7



7 Find the exact value of each expression. Give your answers in simplest surd form where necessary, with a rational denominator.

a b c d

e f g h

8 Show that:a sin 60° cos 30° + cos 60° sin 30° = 1

b cos 45° cos 30° + sin 45°

9 Solve each equation for θ, where 0° < θ < 90°.

a b c

d e f

g h tan θ = 1 i

10 Find the exact value of the pronumeral in each of these. All lengths are in metres.a b c

d e f

11 Find the exact value of the pronumeral in each of these. All lengths are in cm.a b c

sin 30°cos 60°----------------- tan 45°

tan 30°----------------- sin 60°

tan 30°----------------- sin 60°

cos 60°-----------------

tan 60°tan 30°----------------- tan 45°

cos 30°----------------- cos 30°

cos 45°----------------- sin 45°

sin 60°----------------

sin 30° 6 2+4

--------------------=

sin θ 12---= tan θ 3= cos θ 1

2-------=

sin θ 32

-------= cos θ 12---= tan θ 1

3-------=

cos θ 32

-------= sin θ 1

2-------=

45°

x

7

30°

k1060°

t

8

60°

n

8 3 30°

u6

45°

e

9 2

45°

h7 2

30°

8.5

p

60°w6



12 Find the size of the angle θ in each of these. All lengths are in mm.a b c

d e f

13 a Use the exact values to find the height, h cm, of this equilateral triangle.

b Hence, find the exact area of the triangle.

14

a Use the exact values to find the value of x.b Find the value of y.c Hence, find the area of the trapezium.


15 In the rhombus PQRS, the diagonal PR hasa length of cm and ∠SPQ = 120°.a Use the exact values to find the length of

the diagonal SQ.b Hence, find the area of the rhombus.

16 A man walked from S along level ground towards Q, the foot of a vertical cliff PQ. From S, the angle of elevation to the top of the cliff is 30°. After walking to R, he noted that the angle of elevation was then 60°. If the height of the cliff is 120 metres, find the distance RS, without using a calculator.

5 2 5

θ22

11θ

14 3

14

θ

5

10

θ

7 3

7

θ

6 3

9

θ

h cm4 cm 4 cm

4 cm

x cm8 2 cm

5 cm

y cm45°

10 3 cm

P Q

S R

120°10 3

P

Q SR60° 30°

120 m



A bearing is a measure of the direction of one point from another point. There are two types of bearings: compass bearings and true bearings.

■ Compass bearings

A compass bearing is a deviation from northor south and towards east or west. For example, a bearing of S30°W means a deviation of 30° from the south towards the west. The diagramto the right shows the compass bearings of four pints A, B, C, D from a point P.

A bearing such as NE means N45°E. Similarly, NW means N45°W, SE means S45°E and SW means S45°W.

Square area

ABCD is a square and APB is an equilateral triangle.

If AB = 2 units, what is the exact shaded area?

DP

C

A B

TRY THIS

2.8 Bearings Review

P

NE

EW

NW

N

S

SESW

A

B

P

C

EW

N

S

S65°W

S50°E

N72°W

N30°E

50°65°

72°

30°D



■ The 16 point compass

The 16 point compass, or mariner’s compass, was used by mariners to determine directions while at sea. It is still used today in many aspects of everyday life, such as for the wind directions on daily weather reports.

The directions on the 16 point compass are givenas deviations from the four cardinal directions (i.e. N, S, E, W) and towards the four intermediate directions (i.e. NE, NW, SE, SW).

For example, the bearing between north and north-east is north north east (NNE). The bearingbetween north east and east is east north east (ENE), not north east east (NEE).

■ True bearings

A true bearing is a deviation from north, measured in a clockwise direction. Truebearings are written using three digits.For example, a clockwise rotation fromnorth of 52° is written as 052°. The diagramto the right shows the true bearings of four points J, K, L, M from a point P.

■ Opposite bearings

The opposite bearing of B from A is the bearing of A from B. To find the opposite bearing (or any change in direction) it will be necessary to draw a new compass at the end of the ray.

NOTE: (1) Opposite bearings always differ by 180°. That is, the new bearing will be either 180° more than the original bearing, or 180° less.

(2) The reference given after the word ‘FROM’ should be at the centre of the active compass (i.e. the compass in which you are working).

NNNW

NW

WN

WW

WSW

SW

SSWS

SSE

SE

ESE

EE

NE

NE

NNE350 100

2030

4050

6070

80

340

330

320

310

300

290

280

170190 180200210

220

230240250260

160

150

140

130

120110

10090

270

P

M

L

K

J

324°T

060°T

140°T195°T

EW

N

S

36°60°

40°

15°

To find the bearing of A from B given the bearing of B from A:�� draw a compass at B�� mark on this compass the angle from north around to the ray BA�� on the compass with centre A, find the acute angle between BA and the

north–south axis�� use parallel line properties to find the required bearing on the compass with

centre B.



Example 1The bearing of L from M is 162°. Find the bearing of M from L.

Solutioni ∠MLN′ = 18° (Co-interior ∠s, NM || N′L)

ii Bearing of M from L = 360° − 18°= 342°

Example 2Susie drove 12.8 km on a bearing of 228°.How far west did she drive, correct to1 decimal place?

Solution

sin 48° =

∴ x = 12.8 × sin 48°= 9.512253766= 9.5 km (correct to 1 decimal place)

Example 3Daniel walked due north for 3.1 km fromA to B, then turned and walked 4.5 km dueeast to C. Find the bearing of C from A.

Solution

tan θ =

∴ θ = tan−1

� 55°∴ The bearing of C from A is 055°.

EG +S

M

L

EW

N

S

E'W'

N'

S'

162°

18°

EG +S

x12.8----------

EW

N

48°

228°

12.8 km

S

x

EG +S

EW

N

3.1 km

S

B C

A

4.5 km

θ4.53.1-------

4.53.1-------⎝ ⎠

⎛ ⎞



1 i Find the compass bearings from P of the points A, B and C.ii Find the true bearings from P of the points A, B and C.

a b c

2 Use a pair of alternate angles to find the bearings of X from Y, given that the bearing of Yfrom X is:a 030° b 235° c 148° d 303°

3 Find the size of ∠PQR given that:a i the bearing of Q from P is 034° b i the bearing of R from P is 163°

ii the bearing of Q from R is 025° ii the bearing of Q from R is 317°

■ Consolidation

4 Answer the following questions, correct to 1 decimal place.a Albert drove 54 km on a bearing of 065°. How far east was he then from his starting

point?b Nerida flew a light plane 116 km on a bearing of 304°. How far north was she then from

her starting point?c A car travels 115 km due south from L to M, then turns and travels due east to N. If the

bearing of N from L is 133°, find the distance MN.d A cruise ship leaves port and sails due west from P to Q then turns and sails 95 km due

south to R. If the bearing of R from P is 224°, find the distance PQ.

5 Find the following distances, correct to 3 significant figures.a A helicopter flew 26.4 km due north from A to B, then turned and flew due west to C.

If the bearing of C from A is 287°, find the distance AC.b A hiker walked 6.8 km due south from E to F. After a short rest, he then walked due

west to G, which is on a bearing of 216° from E. Find the distance between E and G.

Exercise 2.8

EW

N

50°

15°

66°

P

A

B

C

S

EW

N

S

74°61°

P

AB

C9°

EW

N

S

18°57°

46°P

A

B

C

P R

Q

N N

N

P

R

Q

N

N



c During a cross-country rally, Marcus drove his race car due north from X to Y. He then drove 21.4 km due east to Z, which is on a bearing of 067° from X. Calculate the distance XZ.

d A yacht sailed 11.2 km due east from T, then turned around a buoy at U and sailed due south to V. The bearing of V from T is 138°. How far was the yacht then from its starting point?

6 Give the following bearings as three-figure bearings, correct to the nearest degree.a A woman drove 25 km due east from F to G, then drove 29 km due south to H. Find the

bearing of:i H from F ii F from H

b A man rode due north from I to J, then rode 10 km due west to K, which is 14 km from I. Find the bearing of:i K from I ii I from K

c Marta flew her glider 28 km due south from A to B. She then turned and flew due west to C, which is 36 km from A. Find the bearing of:i C from A ii A from C

d Keiran sailed 19 km due east from R to S, then sailed 22 km due north to T. Find the bearing of:i T from R ii R from T

7 a Alice walked from home (H) to the cinema (C) on a bearing of 043°. After the movie, she walked on a bearing of 133° to a friend’s house (F) 320 m due east of her home.i Show that ∠HCF = 90°.

ii Find the distance between Alice’s home and the cinema, correct to the nearest m.b Cruz drove from home (H) to the beach (B) on a bearing of 254° to pick up his children.

He then drove to the supermarket (S) on a bearing of 344°, which is 9.6 km due west of his home.i Show that ∠HBS = 90°.

ii Find the distance between the beach and the supermarket, correct to the nearest tenth of a km.


8 Two cars A and B left home at the same time. Car A travelled due west at 70 km/h whilst car B travelled due north at 90 km/h. Find, after 2 hours:a the distance between the cars b the bearing of B from A.

9 During a yacht race, a yacht sailed from a harbour (H) due west for 16 nautical miles to buoy X. After sailing around the buoy, the yacht then proceeded to another buoy Y on a course of 288° for 13 nautical miles.a How far west has the yacht sailed from the harbour, correct to 2 decimal places?b How far north has the ship sailed from the harbour, correct to 2 decimal places?c Find the distance between the second buoy (Y) and the harbour, correct to the nearest

nautical mile.d On what bearing must the yacht then sail in order to return directly to the harbour?

10 From a point R, two planes P and Q flying due west and in the same horizontal line are on bearings of 115° and 157° respectively. If P is 200 km from R and Q is 150 km from R, find the distance between the planes, correct to the nearest km.



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ORIENTEERING

Introduction

Orienteering is a sport in which competitors pass through a number of check points called controls, to complete a course aided only by a map and a compass. On the route, orange-and-white control markers are set in the places that correspond to the marked control points on the map. Participants punch their control cards (or use an electronic equivalent) at each control point. Orienteering on foot can be thought of as running navigation. You need to be able to accurately read maps and find direction quickly to be good at it. You also need to be able to make quick decisions and concentrate under stress. There are various types of orienteering including ski, mountain bike and street orienteering but foot orienteering is easily the most popular.

In competitive orienteering the winner is the person who has used the shortest time to visit the control points in the correct order. Fast running alone does not make you a winner. You must also choose the best route between control points and find the markers without wasting unnecessary time. At elite level Australia enters teams in the world championships but it is popular with people of all ages and fitness levels.

FO C U S O N WO R K I N G MA T H E M A T I C A L L Y0 F O C U S O N W 0 R K I N G M A T H E M A T I C A L L Y



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L E A R N I N G A C T I V I T I E S

The baseplate compass

The base plate compass was invented in 1930 in Sweden for the sport of orienteering. It enabled a great saving in time, as competitors could transfer compass readings to maps directly. The diagram below shows the essential features of the compass. It is not possible to explain how it works here. Check with your teacher or the internet. It is not necessary for the learning activity to follow.

An orienteering problem

The map on the next page shows the leg between two control points 17(P) and 18(Q) which competitors must pass through on an orienteering course in the Australian bush. The leg means crossing a wide stream, shallow at the banks but deep enough in the centre to mean getting wet to the waist wading across. North lines on an orienteering map point to magnetic north and are usually spaced 500 m apart. The map is not drawn to scale.

The shortest route from P to Q, a straight line, means traversing mud flats and the sand hills beyond. The terrain marked on the map shows the river bank on the other side is also covered with areas of thick bush and makes a less direct route seem more attractive. The south side of the stream looks easier to navigate, but there is also an area of thick bush to get around. Downstream at R the map shows a flat rock which gives a better view of the area surrounding control point Q.

2

N

S

E

W10 mm 20 30 70

1 I

NC

HE

S1 2

North signIndex line

Direction of travel arrow

Base plate

Magnetic needle

360° dial

Orienting lines



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The team members decide to run downstream to R and cross the river at the sand bank. Estimate the additional distance this route entails.

1 Using the side of the baseplate compass, the distance on the map from P to R is 5.4 cm and the scale on the map is 1:15 000. Make a rough copy of the map in your book and calculate the actual distance from P to R. Show it on your map.

2 The bearing of Q from P is 071° and the bearing of R from P is 120°. Show these angles on your map.

3 The bearing of Q from R is 011° and the bearing of P from R is 282°. Show these angles on your map.

4 Calculate the size of angle PRQ and angle QPR. Show these angles on your map.

5 Given the size of angle PRQ, is it reasonable to estimate the distance RQ by using a trigonometric ratio? What ratio would you choose? Estimate RQ (nearest 10 m).

6 Use a suitable trigonometric ratio to estimate the distance PQ (nearest 10 m).

7 Calculate the extra distance for the route via R. Is the answer reasonable?

C H A L L E N G E A C T I V I T I E S

1 Make a scale drawing of the map and measure the lengths of PR, RQ and PQ. Calculate the extra distance travelled by the team. How does the answer compare with the estimate calculated from trigonometric ratios?

2 If possible learn to use a base plate compass. The class could repeat the exercise above by choosing three distinctive landmarks P, Q and R in the playground, take bearings and measure distance PR with a tape. Why will the angle PRQ need to be approximately 90° to get a good estimate of RQ and PQ using trigonometric ratios?

P

Q

R

NThick bush

Thickbush

Control point 18

Control point 17Rock

River

Mud flats

Sand bank

Diagram not drawn to scale

8


Mathscape 10 extens i onF

OC

US

O

N W

OR

KIN

G M

AT

HE

MA

TIC

AL

LY

68

3 Make an orienteering map using directions from the internet. For example the Orienteering Association of Western Australia has produced an excellent Parkland Mapping Manualwhich schools can download from the Orienteering Australia website<www.orienteering.asn.au>. Click on the links for mapping and then park mapping. Park maps are typically designed to fit A4 or A3 sized paper so that orienteers, schools, community groups and individuals can photocopy them and enjoy an introduction to orienteering in a familiar environment.

L E T ’S C O M M U N I C A T E

In about half a page of writing, explain the sport of orienteering and what mathematical skills are needed to participate.

R E F L E C T I N G

Reflect on the importance of a map and a compass to find direction in the physical world. Why were maps originally made?

E

%

1 Give a meaning for:a compass bearingb angle of elevationc sin A, tan A and cos A

2 Explain the sixteen points of the compasswith a drawing.

3 Read the Macquarie Learners Dictionaryfor bearing:

bearing noun 1. the way you stand or behave: a man of proud bearing 2. a supporting part of machine 3. bearings, direction or position: We lost our bearings in the dark.–phrase 4. have a bearing on, to have a connection or relevance to: This information has no bearing on the problem.

Do you think this explains the mathematical meaning of the word adequately? What is missing?



CH A P T E R RE V I E W

CH

AP

TE

R R

EV

IEW

1 Find without simplifying, the value of each ratio below.

a sin ∠FEG b cos ∠FHEc tan ∠FGE d sin ∠FHEe tan ∠FEH f cos ∠FEG

2 Find the value of

x if .

3 If , draw a right-angled

triangle and hence find the value of sin θ and cos θ.

4 Evaluate ,

correct to 2 decimal places.

5 Find the acute angle θ, correct to the nearest minute.a tan θ = 3.6816 b

6 If tan θ = 1.0729, find the value of sin θ, correct to 4 decimal places.

7 Find the value of each pronumeral, correct to 1 decimal place.a b

c

8 Find the length of the hypotenuse in each of these, correct to 1 decimal place.a

b

9 Find the angle θ, correct to the nearest minute.a

b

c

10 a In ∆ABC, ∠A = 90°, ∠C = 62°45′ and BC = 70 cm. Find the length of AB, correct to 3 significant figures.

b In ∆LMN, ∠M = 90°, ∠L = 73°21′and LM = 36.7 cm. Find the length of LN, correct to the nearest mm.

c In ∆FGH, ∠H = 90°, GH = 19 cm and FH = 10 cm. Find ∠F, correct to the nearest minute.

H FG

E

11 7

2425

30

x20

θcos θ 34---=

tan θ 6572------=

tan 76°19′cos 12°36′ sin 64°10′–--------------------------------------------------------

sin θ 821------=

a

56°12'20

p27°40'

35

z

58°8'26

22°14'

10 cm

h cm

23.8 cm

18°26'

u cm

5

11

θ

25.3

26.8θ

3.7

25.2θ




CH

AP

TE

R R

EV

IEW

11 A building casts a shadow 35 m long on level ground when the altitude of the sun is 51°17′. Find the height of the building, correct to the nearest metre.

12 A wheelchair ramp is inclined at an angle of 15°43′ and has a vertical rise of 1.64 m. Find the length of the ramp, correct to the nearest cm.

13 A surveyor sights the top of a mountain from a distance of 2400 m. The angle of elevation of the peak is 47°29′. Find the height of the mountain, correct to the nearest metre.

14 A man standing on top of a cliff of height 155 m looks down to a boat that is anchored 115 m from the base of the cliff. Find the angle of depression of the boat from the top of the cliff, correct to the nearest minute.

15 A water pipe runs along the slope of a 295 m high hill. The pipe is 372 m long. At what angle is the pipe inclined to the horizontal? Answer correct to the nearest minute.

16 In the isosceles triangle XYZ, YW ⊥ XZ,XY = YZ = 62 cm and ∠XYZ = 104°18′.a Find the altitude YW, correct to the

nearest centimetre.b Find the length of XZ, correct to the

nearest centimetre.

17 An inverted conehas a slant heightof 52 cm and adiameter of 40 cm.Find the verticalangle θ, correct tothe nearest minute.

18 Which one of the following statements is true?

A B

C D

19 Simplify:a cos θ tan θ b

20 Solve each equation for θ, where0° < θ < 90°. Answer correct to the nearest minute.a sin θ = 4 cos θb cos θ = 2 sin θc 5 cos θ = 3 sin θ

21 Find x, if 0° < x° < 90°.a sin x° = cos 70°b cos x° = sin 25°c cos (2x)° = sin 58°d sin (x − 23)° = cos 40°

e

f sin x° = cos x°

22 Simplify:

a b

c cos θ sin (90° − θ)d tan θ sin (90° − θ)

23 Simplify each of these without the use of a calculator. Give your answers in simplest surd form with a rational denominator.a cos 45° cos 60°b sin 60° cos 30°c sin 45° + cos 45°d tan 30° + sin 60°e 1 − 2 sin2 30°f 2 cos2 30° + tan2 60°

g h

Y

ZX W

62 cm

40 cm

52 cm

θ

sin θ cos θtan θ-------------= cos θ tan θ

sin θ------------=

tan θ sin θcos θ------------= tan θ cos θ

sin θ------------=

sin θtan θ------------

sinx2---⎝ ⎠

⎛ ⎞ ° cos 78°=

cos 90° θ–( )sin θ

-------------------------------- cos 90° θ–( )cos θ

--------------------------------

tan 45°sin 30°----------------- cos 45°

sin 60°-----------------




CH

AP

TE

R R

EV

IEW

24 Find the exact value of x.a b

25 Find θ, without the useof a calculator.

26 Find i the compass bearing andii the truebearingof eachpointfrom P.

a A b B c C d D

27 Find the bearing of X from Y, given that the bearing of Y from X is:a 112° b 237°

28 Find the size of∠ABC, given that the bearing of Bfrom A is 235° and the bearing of B from C is 217°.

29 a Amber drove 126 km on a bearing of 137°. How far south did she drive? Answer correct to 1 decimal place.

b Ken rode his horse 8.7 km on a bearing of 295°. How far west did he ride? Answer correct to 1 decimal place.

30 Answer each of the following, correct to 4 significant figures.a A ship leaves port and sails

60 nautical miles due east from L to M, then turns and sails due north to N. If the bearing of N from L is N24°E, find the distance MN.

b Beaumont Hills is due west of Coleville. Coleville is 11 km due north of Allentown. The bearing of Allentown from Beaumont Hills is 136°. Find the distance between Beaumont Hills and Allentown.

31 Answer each of the following as a true bearing, correct to the nearest degree.a A ferry sailed 12 km due south from

T to U, then turned and sailed 19 km due east to V. Find the bearing of:i V from T ii T from V

b Peter drove 94 km due north from Cto D, then turned and drove 52 km due west to E. Find the bearing of:i E from C ii C from E

32 Xian walked 250 m from I to J on a bearing of 043°. He then turned and walked on a bearing of 133° to K, which is due east of I.a Show that ∠IJK = 90°.b Find the distance JK, correct to the

nearest metre.

33 Angus drove from P to Q on a bearing of 211°. He then turned and drove 48 km on a bearing of 301° to R, which is due west of P.a Show that ∠PQR = 90°.b Find the distance PR, correct to the

nearest kilometre.

x m

2 3 m60°

x m30°

6 m

7 2 7

θ

55°

9°

13°EW

S

N

BC

D

A

48° P

N

N

N

B

A

C


chap2

Documents

cosine ratio cos

sine ratio sin

cos psq5

cos adjacenthypotenuse

tangent ratio tan

tan oppositeadjacent

cosine ratios

bearing of b